Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y + 2}{y^2 - 13y + 36} \div \dfrac{-10y - 20}{y^2 - 9y} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{y + 2}{y^2 - 13y + 36} \times \dfrac{y^2 - 9y}{-10y - 20} $ First factor the quadratic. $z = \dfrac{y + 2}{(y - 9)(y - 4)} \times \dfrac{y^2 - 9y}{-10y - 20} $ Then factor out any other terms. $z = \dfrac{y + 2}{(y - 9)(y - 4)} \times \dfrac{y(y - 9)}{-10(y + 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (y + 2) \times y(y - 9) } { (y - 9)(y - 4) \times -10(y + 2) } $ $z = \dfrac{ y(y + 2)(y - 9)}{ -10(y - 9)(y - 4)(y + 2)} $ Notice that $(y + 2)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ y(y + 2)\cancel{(y - 9)}}{ -10\cancel{(y - 9)}(y - 4)(y + 2)} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $z = \dfrac{ y\cancel{(y + 2)}\cancel{(y - 9)}}{ -10\cancel{(y - 9)}(y - 4)\cancel{(y + 2)}} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $z = \dfrac{y}{-10(y - 4)} $ $z = \dfrac{-y}{10(y - 4)} ; \space y \neq 9 ; \space y \neq -2 $